Solutions to problem set 3 (141A Sp07)
1
1. (ISSP 3.3) Basis of two unalike atoms.
At
k
=
π/
a
, equation (20) becomes
2

ω
2
M
1
C
!
u
=
0
2

ω
2
M
2
C
!
v
=
0
Note that the
u
and
v
sublattices are decoupled at this value of
k
; u can change without
changing
v
, and vice versa. There are two solutions to the above equations:
u
=
0,
ω
2
=
2
C
/
M
2
and
v
arbitrary; and
v
=
0,
ω
2
=
2
C
/
M
1
and
u
arbitrary.
2. (ISSP 3.4) Kohn anomaly. Equation (16a) gives
ω
2
=
2
A
M
∞
X
p
=
1
sin
pk
0
a
pa
(1

cos(
pka
))
=
2
A
Ma
∞
X
p
=
1
sin
pk
0
a
p

1
2
sin
p
(
k
0
+
k
)
a
p

1
2
sin
p
(
k
0

k
)
a
p
!
Gradshteyn and Ryzhik (
Table of integrals, series and products
)1.441 gives
∞
X
p
=
1
sin
α
p
p
=
π

α
2
0
< α <
2
π
(you can prove this by writing the sin as the sum of two exponentials, and then using the
taylor expansion of
log
(1
+
x
).)
So for
k
0
>
k
, we have
ω
2
=
2
A
Ma
∞
X
p
=
1
sin
pk
0
a
p

1
2
sin
p
(
k
0
+
k
)
a
p

1
2
sin
p
(
k
0

k
)
a
p
!
=
2
A
Ma
π

k
0
a
2

1
2
π

(
k
0
+
k
)
a
2

1
2
π

(
k
0

k
)
a
2
!
=
0
Now consider the case where
k
0
<
k
. In order to apply the formula from the tables we need
the argument of each sin function to be positive. Then
ω
2
=
2
A
Ma
∞
X
p
=
1
sin
pk
0
a
p

1
2
sin
p
(
k
0
+
k
)
a
p

1
2
sin
p
(
k
0

k
)
a
p
!
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 Spring '08
 SOUZA
 Derivative, Solid State Physics, Sin, C2 Greyhound, Massachusetts Route 2A, nearest neighbour coupling

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